Average Error: 41.1 → 0.6
Time: 2.8s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.9847889305445352:\\ \;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\ \mathbf{else}:\\ \;\;\;\;e^{x} \cdot \left(\left(\frac{1}{12} \cdot x + \frac{1}{x}\right) - \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.9847889305445352:\\
\;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\

\mathbf{else}:\\
\;\;\;\;e^{x} \cdot \left(\left(\frac{1}{12} \cdot x + \frac{1}{x}\right) - \frac{1}{2}\right)\\

\end{array}
double code(double x) {
	return ((double) (((double) exp(x)) / ((double) (((double) exp(x)) - 1.0))));
}
↓
double code(double x) {
	double VAR;
	if ((((double) exp(x)) <= 0.9847889305445352)) {
		VAR = ((double) (((double) exp(x)) / ((double) log(((double) exp(((double) (((double) exp(x)) - 1.0))))))));
	} else {
		VAR = ((double) (((double) exp(x)) * ((double) (((double) (((double) (0.08333333333333333 * x)) + ((double) (1.0 / x)))) - 0.5))));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.1
Target40.7
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9847889305445352

    1. Initial program 0.0

      \[\frac{e^{x}}{e^{x} - 1}\]
    2. Using strategy rm

    3. Applied add-log-exp0.0

      \[\leadsto \frac{e^{x}}{e^{x} - \color{blue}{\log \left(e^{1}\right)}}\]

    4. Applied add-log-exp0.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\log \left(e^{e^{x}}\right)} - \log \left(e^{1}\right)}\]

    5. Applied diff-log0.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\log \left(\frac{e^{e^{x}}}{e^{1}}\right)}}\]

    6. Simplified0.0

      \[\leadsto \frac{e^{x}}{\log \color{blue}{\left(e^{e^{x} - 1}\right)}}\]

    if 0.9847889305445352 < (exp x)

    1. Initial program 61.7

      \[\frac{e^{x}}{e^{x} - 1}\]

    2. Taylor expanded around 0 1.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\frac{1}{2} \cdot {x}^{2} + \left(\frac{1}{6} \cdot {x}^{3} + x\right)}}\]

    3. Simplified1.0

      \[\leadsto \frac{e^{x}}{\color{blue}{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}}\]
    4. Using strategy rm

    5. Applied div-inv1.0

      \[\leadsto \color{blue}{e^{x} \cdot \frac{1}{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}}\]

    6. Taylor expanded around 0 0.9

      \[\leadsto e^{x} \cdot \color{blue}{\left(\left(\frac{1}{12} \cdot x + \frac{1}{x}\right) - \frac{1}{2}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.9847889305445352:\\ \;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\ \mathbf{else}:\\ \;\;\;\;e^{x} \cdot \left(\left(\frac{1}{12} \cdot x + \frac{1}{x}\right) - \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020124 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

  :herbie-target
  (/ 1.0 (- 1.0 (exp (- x))))

  (/ (exp x) (- (exp x) 1.0)))